Is Calkin algebra $B(H)/\mathcal K(H)$ isomorphic to some subalgebra of $B(H)$?

Question

$B(H)$ is the operator algebra on separable Hilbert space. $\mathcal K(H)$ is the compact operator algebra.

I have no idea on this. Could someone please give me a hint?

Answer 1

It is easy to show that the Calking algebra has an uncountable family of pairwise orthogonal projections, so it cannot embed in $B(H)$.

The way to show this is to construct an uncountable family$\{A_t\}$ of infinite subsets of $\mathbb N$ such that, for all $s\ne t$, the set $A_t\cap A_s$ is finite. Then, given a countable family $\{E_n\}$ of pairwise orthogonal rank-one projections in $B(H)$, one forms projections $$ P_t=\sum_{n\in A_t}E_n. $$ Then $ \displaystyle P_tP_s=\sum_{n\in A_t\cap A_s}E_n $ is finite-rank, so in the Calkin $$ \pi(P_t)\pi(P_s)=\pi(P_tP_s)=0. $$

Answer 2

From the Wikipedia page on the Calkin Algebra:

"As a C*-algebra, the Calkin algebra is not isomorphic to an algebra of operators on a separable Hilbert space"